Question: Let $h$ be a vector-valued function defined by $h(t)=(3t^2+t,2\cdot3^t)$. Find $h'(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $(6t+1,2\ln(3)\cdot3^t)$ (Choice B) B $(6t,2\cdot3^t)$ (Choice C) C $(6t^2,2\ln(2)\cdot3^t)$ (Choice D) D $6t-2\cdot3^t$
Solution: $h$ is a vector-valued function. This means it takes one number as an input $(t)$, but it outputs two numbers as a two-dimensional vector. Finding the derivative of a vector-valued function is pretty straightforward. Suppose a vector-valued function is defined as $u(t)=(v(t),w(t))$, then its derivative is the vector-valued function $u'(t)=(v'(t),w'(t))$. In other words, the derivative is found by differentiating each of the expressions in the function's output vector. Recall that $h(t)=(3t^2+t,2\cdot3^t)$. Let's differentiate the first expression: $\begin{aligned}\dfrac{d}{dt}(3t^2+t)&=2\cdot3t+1 \\\\&=6t+1\end{aligned}$ Let's differentiate the second expression: $\dfrac{d}{dt}(2\cdot3^t)=2\ln(3)\cdot3^t$ Now let's put everything together: $\begin{aligned} h'(t)&=\left(\dfrac{d}{dt}(3t^2+t),\dfrac{d}{dt}(2\cdot3^t)\right) \\\\ &=(6t+1,2\ln(3)\cdot3^t) \end{aligned}$ In conclusion, $h'(t)=(6t+1,2\ln(3)\cdot3^t)$.